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Beliebt Rechnen >

integral of 1/(1+sqrt(4x^2+1))

Lösung

\int \frac{1}{1 + 4 x ^{2} + 1} d x

Lösung

\frac{1}{2} (ln tan (\frac{arctan ( 2 x )}{2}) + 1 - ln tan (\frac{arctan ( 2 x )}{2}) - 1 - tan (\frac{arctan ( 2 x )}{2})) + C

Schritte zur Lösung

\int \frac{1}{1 + 4 x ^{2} + 1} d x

Trigonometrische Substitution anwenden

= \int \frac{sec ^{2} ( u )}{2 ( 1 + sec ( u ) )} d u

Entferne die Konstante:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot \int \frac{sec ^{2} ( u )}{1 + sec ( u )} d u

Vereinfache

\frac{sec ^{2} ( u )}{1 + sec ( u )} : \frac{1}{cos ( u ) + cos ^{2} ( u )}

= \frac{1}{2} \cdot \int \frac{1}{cos ( u ) + cos ^{2} ( u )} d u

Wende U-Substitution an

= \frac{1}{2} \cdot \int \frac{v ^{2} + 1}{- v ^{2} + 1} d v

\frac{v ^{2} + 1}{- v ^{2} + 1} = \frac{2}{- v ^{2} + 1} - 1

= \frac{1}{2} \cdot \int \frac{2}{- v ^{2} + 1} - 1 d v

Wende die Summenregel an:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{2} (\int \frac{2}{- v ^{2} + 1} d v - \int 1 d v)

\int \frac{2}{- v ^{2} + 1} d v = ln ∣ v + 1 ∣ - ln ∣ v - 1 ∣

\int 1 d v = v

= \frac{1}{2} (ln ∣ v + 1 ∣ - ln ∣ v - 1 ∣ - v)

Ersetze zurück

= \frac{1}{2} (ln tan (\frac{arctan ( 2 x )}{2}) + 1 - ln tan (\frac{arctan ( 2 x )}{2}) - 1 - tan (\frac{arctan ( 2 x )}{2}))

Füge eine Konstante zur Lösung hinzu

= \frac{1}{2} (ln tan (\frac{arctan ( 2 x )}{2}) + 1 - ln tan (\frac{arctan ( 2 x )}{2}) - 1 - tan (\frac{arctan ( 2 x )}{2})) + C

Graph

Beliebte Beispiele

integral of x/(4sqrt(9-x^2))

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\int \frac{1}{( 1 - \frac{2}{9} x ) ^{3}} d x

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\int \frac{d}{d x} (sin (x)) d x